Upload
imam-genji
View
120
Download
6
Embed Size (px)
Citation preview
Leibniz Rule for Differentiating Products
Formula to find a high order derivative of product.Example:
d9
dx9(x sin x) = x
d9
dx9(sin x) + 9
d
dx(x)
d8
dx8(sin x)
+9 ⋅ 82!
d2
dx2(x)
d7
dx7(sin x) + ⋅ ⋅ ⋅
= xd9
dx9(sin x) + 9
d8
dx8(sin x)
+9 ⋅ 82!⋅ 0 ⋅ d7
dx7(sin x)
= x cos x + 9 sin x
Leibniz Rule for Differentiating Products
Binomial expansion
(a + b)9 = a0b9 + 9ab8 +9 ⋅ 82!
a2b7 + ⋅ ⋅ ⋅
Exercise: Try to differentiate this
d10
dx10(xex)
Rodrigues’ Formula
Another way of finding Legendre Polynomials
Pl(x) =1
2l l!
dl
dx l(x2 − 1)l
Rodrigues’ Formula: Cont’dExample: when l = 0
P0(x) =1
200!
d0
dx0(x2 − 1)0
= (x2 − 1)0 = 1
When l = 1
P1(x) =1
211!
d1
dx1(x2 − 1)1
=1
2(2x) = x
When l = 2
P2(x) =1
222!
d2
dx2(x2 − 1)2
=1
8(12x2 − 4) =
1
2(3x2 − 1)
Rodrigues’ Formula: Proof
Let v = (x2 − 1)l . Differentiate once
dv
dx= l(x2 − 1)l−1(2x)
(x2 − 1)dv
dx= l(x2 − 1)(2x) = l ⋅ v ⋅ 2x
(x2 − 1)dv
dx= 2xlv (1)
Rodrigues’ Formula: Proof Cont’d
Differentiate Eqn. (1) l + 1 times using Leibniz rule
dl+1
dx l+1
((x2 − 1)
dv
dx
)=
dl+1
dx l+1(2xlv)
(x2 − 1)dl+2v
dx l+2+ (l + 1)(2x)
dl+1v
dx l+1+
(l + 1)l
2!⋅ 2 ⋅ d
lv
dx l
= 2lxdl+1v
dx l+1+ 2l(l + 1)
dlv
dx l(2)
Rodrigues’ Formula: Proof Cont’d
Rearrange Eqn. (2), we get
(1− x2)dl+2v
dx l+2− 2x
dl+1v
dx l+1+ l(l + 1)
dlv
dx l= 0
(1− x2)
(dlv
dx l
)′′
− 2x
(dlv
dx l
)′
+ l(l + 1)dlv
dx l= 0
Which show that the Rodrigues’ formula satisfied the Legendreequation.
Rodrigues’ Formula: Proof Cont’d
Here we see that
C ⋅ dlv
dx l= Pl(x)
The derivative for x2n, when v = (x2 − 1)l = (x + 1)l(x − 1)l is
dlv
dx l= (x + 1)l
dl
dx l+l C1 ⋅ l(x + 1)l−1 ⋅ dl−1
dx l−1(x − 1)l
+ ⋅ ⋅ ⋅+ (x − 1)ldl
dx l(x + 1)l = 0
When x = 1,dlv
dx l= 2l ⋅ l!
Rodrigues’ Formula: Proof Cont’d
Thus, when x = 1
C ⋅ dlv
dx l= Pl(1) = 1
Which give C as
C =1
2l ⋅ l!So,
Pl(x) =1
2l ⋅ l!dlv
dx l
=1
2l ⋅ l!dl
dx l(x2 − 1)l
Proved
Generating Function for Legendre Polynomials
Many properties of Legendre Polynomials can be derive by usingGenerating function
Φ(x , h) =1√
1− 2xh + h2, ∣h∣ < 1, (3)
Generating Function for Legendre Polynomials: Cont’d
Maclaurin series for (1− y)−1/2 for −1 < y < 1 is
1√1− y
= 1 +1
2y +
3
8y2 +
15
48y3 +
105
384y4 +
945
3840y5 + ⋅ ⋅ ⋅
Setting y = 2xh − h2, we obtain
1√1− 2xh − h2
= 1 +1
2(2xh − h2) +
3
8(2xh − h2)2
+15
48(2xh − h2)3 +
105
384(2xh − h2)4
+945
3840(2xh − h2)5 + ⋅ ⋅ ⋅
Generating Function for Legendre Polynomials: Cont’d
Expand the previous equation and collecting h term, we get
1√1− 2xh − h2
= 1 + xh +
(−1
2+
3
2
)h2 +
(−3
2x +
5
2x3)h3
+
(3
8− 15
4x2 +
35
8x4)h4
+
(15
8x − 35
4+
63
8x5)h5 + ⋅ ⋅ ⋅
= P0(x) + P1(x)h + P2(x)h2 + P3(x)h3
+ P4(x)h4 + P5(x)h5 + ⋅ ⋅ ⋅
=∞∑l=0
hlPl(x)
Generating Function for Legendre Polynomials: Cont’d
When x = 1
Φ(1, h) = (1− 2h + h2)−1/2
=1
1− h
= 1 + h + h2 + ⋅≡ P0(1) + P1(1)h + P2(1)h2 + ⋅ ⋅ ⋅
Thus Pl(1) = 1.
Generating Function for Legendre Polynomials: Cont’d
To show that generating function satisfy Legendre equation, weuse this equation
(1− x2)∂2Φ
∂x2− 2x
∂Φ
∂x+ h
∂2
∂h2(hΦ) = 0
Where
∂Φ
∂x=
h
(1− 2xh + h2)3/2
∂2Φ
∂x2=
3h2
(1− 2xh + h2)5/2
∂2
∂h2(hΦ) = − −2x + hx2 − 2xh2 + 3h
(−1 + 2xh − h2)√
1− 2xh + h2
Generating Function for Legendre Polynomials: Cont’d
Inserting
Φ(x , h) =∞∑l=0
hlPl(x)
into the previous equation, we get
(1− x2)∞∑l=0
hlP′′l (x)− 2x
∞∑l=0
hlP′l (x) +
∞∑l=0
l(l + 1)hlPl(x) = 0
∞∑l=0
hl(
(1− x2)P′′l (x)− 2xP
′l (x) + l(l + 1)Pl(x)
)= 0
Which is Legendre Equations.
Recursion Relations for Legendre Polynomials
1. lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x)
2. xP′l (x)− P
′l−1(x) = lPl(x)
3. P′l (x)− xP
′l−1(x) = lPl−1(x)
4. (1− x2)P′l (x) = lPl−1(x)− lxPl(x)
5. (2l + 1)Pl(x) = P′l+1(x)− P
′l−1(x)
6. (1− x2)P′l−1(x) = lxPl−1(x)− lPl(x)
Proof for lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x)
Given
Φ(x , h) = (1− 2xh + h2)−1/2 =∞∑l=0
hlPl(x)
We differentiate with respect to h,
∂Φ
∂h= − 1
2 3√
1− 2xh + h2(−2x + 2h)
(1− 2xh + h2)∂Φ
∂h= (x − h)Φ
Or
(1− 2xh + h2)∞∑l=1
lhl−1Pl(x) = (x − h)∞∑l=0
hlPl(x)
Proof for lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x):Cont’d
Expanding the previous equation, we get,
∞∑l=1
lhl−1Pl(x)− 2x∞∑l=1
lhlPl(x) +∞∑l=1
lhl+1Pl(x)
= x∞∑l=0
hlPl(x)−∞∑l=0
hl+1Pl(x)
Adjusting h indices so that all become hl−1, we get
∞∑l=1
lhl−1Pl(x)− 2x∞∑l=2
(l − 1)hl−1Pl−1(x) +∞∑l=3
(l − 2)hl−1Pl−2(x)
= x∞∑l=1
hl−1Pl−1(x)−∞∑l=2
hl−1Pl−2(x)
Proof for lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x):Cont’d
We can write the equation from the previous slide as
lPl(x)− 2x(l − 1)Pl−1(x) + (l − 2)Pl−2(x) = xPl−1(x)− Pl−2(x)
When we rearrange we will get the first recursion formula.
lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x)
Proof for xP′
l (x)− P′
l−1(x) = lPl(x)
Given
Φ(x , h) = (1− 2xh + h2)−1/2 =∞∑l=0
hlPl(x)
Differentiate with respect to h, we get
∂Φ
∂h=
(x − h)3√
1− 2xh + h2=∞∑l=1
lhl−1Pl(x) (4)
Differentiate with respect to x , we get
∂Φ
∂x= − 1
2 3√
1− 2xh + h2(−2h)
=h
3√
1− 2xh + h2=∞∑l=0
hlP′l (x) (5)
Proof for xP′
l (x)− P′
l−1(x) = lPl(x): Cont’d
Dividing Eqn. (4) with Eqn. (5), we get
x − h
h=
∑∞l=1 lh
l−1Pl(x)∑∞l=0 h
lP′l (x)
(x − h)∞∑l=0
hlP′l (x) = x
∞∑l=0
hlP′l (x)−
∞∑l=0
hl+1P′l (x)
=∞∑l=1
lhlPl(x)
Proof for xP′
l (x)− P′
l−1(x) = lPl(x): Cont’d
Adjusting h indices so that all become hl , we get
x∞∑l=0
hlP′l (x)−
∞∑l=0
hl+1P′l (x)
= x∞∑l=0
hlP′l (x)−
∞∑l=1
hlP′l−1(x)
=∞∑l=1
lhlPl(x)
So we get the second recursion relation
xP′l (x)− P
′l−1(x) = lPl(x)
Proof for P′
l (x)− xP′
l−1(x) = lPl−1(x)
The first recursion relation is
lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x)
Differentiate with respect to x , we get
lP′l = (2l − 1)Pl−1 + (2l − 1)xP
′l−1 − (l − 1)P
′l−2
Rearrange, using second recursion relation, we get
l [P′l − xP
′l−1]− (l − 1)[xP
′l−1 − P
′l−2]
= l [P′l − xP
′l−1]− (l − 1)[(l − 1)Pl−1]
= (2l − 1)Pl−1
Proof for P′
l (x)− xP′
l−1(x) = lPl−1(x): Cont’d
Then
l [P′l − xP
′l−1] = (l − 1)(l − 1)Pl−1 + (2l − 1)Pl−1
= l2Pl−1
So we get the third recursion relation.
P′l − xP
′l−1 = lPl−1
Proof for (1− x2)P′
l (x) = lPl−1(x)− lxPl(x)
Using
P′l (x)− xP
′l−1(x) = lPl−1(x) (6)
xP′l (x)− P
′l−1(x) = lPl(x) (7)
Multiply Eqn. (7) with x and subtracting from Eqn. (6), we get
(1− x2)P′l (x) = lPl−1(x)− lxPl(x)
Proof for (2l + 1)Pl(x) = P′
l+1(x)− P′
l−1(x)
Using first recursion relation
lPl(x) = (2l − 1)xPl−1(x)− (l − 1)Pl−2(x)
Replacing l with l + 1, we get
(l + 1)Pl+1 = (2(l + 1)− 1)xPl − (l + 1− 1)Pl−1
= (2l + 1)xPl − lPl−1
Differentiate with respect to x , we get
(l + 1)P′l+1 = (2l + 1)Pl + (2l + 1)xP
′l − lP
′l−1
Proof for (2l + 1)Pl(x) = P′
l+1(x)− P′
l−1(x): Cont’d
Inserting the second recursion formula xP′l − P
′l−1 = lPl into the
previous equation, we get
(l + 1)P′l+1 = (2l + 1)Pl + (2l + 1)(lPl + P
′l−1)− lP
′l−1
(l + 1)P′l+1 − (l + 1)P
′l−1 = (2l + 1)(1 + l)Pl
So we getP
′l+1 − P
′l−1 = (2l + 1)Pl
Proof for (1− x2)P′
l−1(x) = lxPl−1 − lPl(x)
Try to prove this recursion relation on your own.